Rectangle hyperbola

Jul 2009
338
14
Singapore
The tangent at the point on the curve \(\displaystyle x=ct\) , \(\displaystyle y=\frac{c}{t}\) meets the X and y axes at P and Q respectively. The normal at the point meets the line y=x and y=-x at R and S respectively. Prove that PRQS is a rhombus unless \(\displaystyle t^2=1\)

I found
\(\displaystyle P(2ct,0)\) , \(\displaystyle Q(0,\frac{2c}{t})\)
and \(\displaystyle R(\frac{c(t^2+1)}{t},\frac{c(t^2+1)}{t})\)
\(\displaystyle S(\frac{c(t^2-1)}{t},-\frac{c(t^2-1)}{t})\)

I tried the gradients of the diagonals PQ and RS are perpendicular, and got no restrictions on \(\displaystyle t^2\) I also tried finding the lengths of all the sides but they are all the same \(\displaystyle \frac{c}{t}\sqrt{t^4+1}\) so i cannot prove that PRQS is a rhombus unless \(\displaystyle t^2=1\).
Thanks
 
Jun 2009
806
275
At t^2 = 1, S co-ordinates are (0, 0).

Normal at any point on the hyperbola will not pass through the origin, unless it is drawn at the vertex. In that case, R and S will coincide and tangent at P will not cut y-axis.
 
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Jul 2009
338
14
Singapore
is there any like mathematical proof of this as in like from equations and such?
thanks